March 2010
SOME REMARKS ON ALMOST LINDEL ¨OF SPACES AND WEAKLY LINDEL ¨OF SPACES
Yan-Kui Song and Yun-Yun Zhang
Abstract. A spaceX isalmost Lindel¨of (weakly Lindel¨of) if for every open coverUofX, there exists a countable subsetVofUsuch thatS
{V :V ∈ V}=X(respectively,S
V=X). In this paper, we investigate the relationships among almost Lindel¨of spaces, weakly Lindel¨of spaces and Lindel¨of spaces, and also study topological properties of almost Lindel¨of spaces and weakly Lindel¨of spaces.
1. Introduction
By a space we mean a topological space. Let us recall that a space X is Lindel¨of if every open cover ofX has a countable subcover. As a generalization of Lindel¨ofness, Willard and Mathur [8] defined a space X to be almost Lindel¨of if for every open cover U of X, there exists a countable subsetV ofU such that S{V :V ∈ V}=X. Frolik [4] defined a spaceX to beweakly Lindel¨of if for every open cover U of X, there exists a countable subset V of U such that S
V = X.
Clearly, every Lindel¨of space is almost Lindel¨of and every almost Lindel¨of space is weakly Lindel¨of, but the converses do not hold (see Examples 2.2 and 2.3). On the study of almost Lindel¨of spaces and weakly Lindel¨of spaces, the readers can see the references [1, 2, 4, 5, 6].
The purpose of this paper is to investigate the relationships among almost Lindel¨of spaces, weakly Lindel¨of spaces and Lindel¨of spaces, and also to study topological properties of almost Lindel¨of spaces and weakly Lindel¨of spaces.
Recall that theextent e(X) of a spaceXis the smallest cardinal numberκsuch that the cardinality of every discrete closed subset ofX is not greater thanκ. The cardinality of a setAis denoted by|A|. Letω be the first infinite cardinal,ω1 the first uncountable cardinal and cthe cardinality of the set of all real numbers. As usual, a cardinal is the initial ordinal and an ordinal is the set of smaller ordinals.
2010 AMS Subject Classification: 54D20, 54E18.
Keywords and phrases: Lindel¨of; almost Lindel¨of; weakly Lindel¨of.
The author acknowledges support from the National Science Foundation of Jiangsu Higher Education Institutions of China (Grant No. 07KJB110055).
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For a cardinalκ,cf(κ) denotes the cofinalityκ. Every cardinal is often viewed as a space with the usual order topology. Other terms and symbols that we do not define follow [3].
2. Some examples
In this section, we give some examples showing the relationship among almost Lindel¨of spaces, weakly Lindel¨of spaces and Lindel¨of spaces. First, we give a well- known result for the sake of completeness.
Proposition 2.1. IfX is a regular almost Lindel¨of space, then X Lindel¨of.
In the following, we give an example showing that Proposition 2.1 is not true for Urysohn spaces.
Example 2.2. There exists an Urysohn almost Lindel¨of spaceX which is not Lindel¨of.
Proof. Let
A={aα:α < ω1}, B ={bi:i∈ω}, Y ={haα, bii:α < ω1, i∈ω}
and
X =Y ∪A∪ {a} where a /∈Y ∪A.
We topologize X as follows: every point ofY is isolated; a basic neighborhood of aα∈Afor eachα < ω1 takes the form
Uaα(i) ={aα} ∪ {haα, bji:α < ω1, j≥i}wherei∈ω and a basic neighborhood ofatakes the form
Ua(α) ={a} ∪[
{haβ, bii:β > α, i∈ω}}whereα < ω1.
Clearly, X is a Urysohn space. Moreover X is not regular, since the pointa can not be separated from the closed set {aα : α < ω1}. Since {aα : α < ω1} is an uncountable discrete closet set ofX, thenX is not Lindel¨of.
We show thatX is almost Lindel¨of. LetU be any open cover ofX. Then there exists someUa ∈ U such that a∈ Ua. By the definition of topology of X, there exists aβ < ω1 such thatUa(β)⊆Ua, then
{aα:α > β} ∪ {a} ∪ {haα, bii:α > β, i∈ω} ⊆Ua.
It is not difficult to see that X \Ua is at most countable, so there exists a countable subset V of U such that X\Ua ⊆S
V. If we putV ={Ua} ∪ V, then V is a countable subfamily ofU such thatX ={V :V ∈ V}, which completes the proof.
For a Tychonoff spaceX, letβXdenote the ˇCech-Stone compactification ofX.
Example 2.3. There exists a Tychonoff weakly Lindel¨of space which is not almost Lindel¨of.
Proof. LetDbe a discrete space of cardinality ω1, let X= (βD×(ω+ 1))\((βD\D)× {ω}) be the subspace of the product ofβDandω+ 1.
We show that X is weakly Lindel¨of. Let U be any open cover of X. Since βD×ω is aσ-compact dense subset of X, then there exists a countable subsetV ofU such that βD×ω⊆S
V, henceX =S
V, sinceβD×ω is a dense subset of X, which shows thatX is weakly Lindel¨of.
Next, we show thatXis not almost Lindel¨of. Since|D|=ω1, we can enumerate Das {dα:α < ω1}. For eachα < ω1, letUα={dα} ×(ω+ 1). For eachn∈ω, let Vn=βD× {n}. Let us consider the open cover
U ={Uα:α < ω1} ∪ {Vn:n∈ω}
ofX. It is not difficult to see thatS V =S
{V :V ∈ V}for each a countable subset V ofU. LetV be any countable subset of U and letα0 = sup{α:Uα∈ V}. Then α0 < ω1,sinceV is countable. If we pick α0 > α0, then hdα0, ωi ∈ {V/ : V ∈ V}, sinceUα0 is the only element ofU containing hdα0, ωi and S
V =S
{V : V ∈ V}, which completes the proof.
If we takeDof arbitrarily big cardinality instead ofω1in the proof in Example 2.3, we easily get the following result.
Proposition 2.4. For every infinite cardinal κ, there exists a Tychonoff weakly Lindel¨of spaceX such that e(X)≥κ.
Remark 2.1. F. Cammaroto and G. Santoro [2] also constructed an example showing that there exists a Tychonoff weakly Lindel¨of space that is not almost Lindel¨of (see Example 3.11 [2]). Example 2.3 is simpler than their construction.
Remark 2.2. As one of the referees observed, it is easy to see that every CCC space is weakly Lindel¨of, so everyCCC, non Lindel¨of Tychonoff space (for example, aP
-product in 2κ) would work as such an example. However we include Example 2.3 here, since we use it later in the text.
It is well known that the extent of a Lindel¨of space is countable. However, similar to the argument from Example 2.2, we can prove the following proposition showing that the extent of a Urysohn almost Lindel¨of space can be arbitrarily big.
Proposition 2.5. For every infinite cardinalκ, there exists a Urysohn almost Lindel¨of spaceX such that e(X)≥κ.
3. Behavior with respect to products, images and subspaces From Example 2.2, it is not difficult to see that the closed subset of a Urysohn almost Lindel¨of space need not be almost Lindel¨of. The following example shows that a regular closed subspace of a Urysohn almost Lindel¨of spaces need not be almost Lindel¨of.
Example 3.1. There exist a Urysohn almost Lindel¨of spaceXhaving a regular closed subset which is not almost Lindel¨of.
Proof. LetS1 be the space X from Example 2.2 and let S2 be the space X from Example 2.3.
We assume that S1∩S2 =∅. Since|D| =ω1, we can enumerate D as {dα : α < ω1}. Letϕ:D× {ω} →Abe a bijection defined by
ϕ(hdα, ωi) =aαfor eachα < ω1.
LetX be the quotient space obtained from the discrete sumS1⊕S2 by identifying hdα, ωiwithϕ(hdα, ωi) for eachα < ω1. Letπ:S1⊕S2→X be the quotient map.
Let Y = π(S2). Then, Y is not almost Lindel¨of in X since it is homeomorphic toS2.
Now, we showX is almost Lindel¨of. LetU be an open cover ofX. Sinceπ(S1) is almost Lindel¨of, then there exists a countable subfamilyV0ofUsuch thatπ(S1)⊆ S{V :V ∈ V0}; on the other hand, for eachn∈ω, sinceπ(βD× {n}) is a compact subset ofX, there exists a finite subfamilyVnofU such thatπ(βD× {n})⊆S
Vn. If we put V=V0∪S
{Vn :n∈ω}, thenV is a countable subfamily ofU such that X =S
{V :V ∈ V}, which shows thatX is almost Lindel¨of.
From Example 2.3, it is not difficult to see that the closed subset of a Tychonoff weakly Lindel¨of space need not be weakly Lindel¨of. However we have the following positive result.
Proposition 3.2. Every regular closed subset of a weakly Lindel¨of space X is weakly Lindel¨of.
Proof. LetX be a weakly Lindel¨of space and letF be a regular closed subset ofX. LetU be an open cover ofF. For eachU ∈ U, there exists an open subset VU in X such that VU ∩F = U. Then {VU : U ∈ U} ∪ {X \F} is an open cover of X. Hence there exists a countable subsetV of {VU :U ∈ U} ∪ {X \F} such that X = S
V, sinceX is weakly Lindel¨of. Let W = V \ {X \F}. Then IntF ⊆ S
W. Hence F = IntF ⊆ S
W, since F is a regular closed subset of X. Thus F = F ∩S
W = clF(F ∩(S
W)) = clF(∪{F ∩W : W ∈ W}. Since {F ∩W :W ∈ W}is a countable subset ofU and F = clF(S
{F∩W :W ∈ W}), thenF is weakly Lindel¨of, which completes the proof.
The following positive results are obvious.
Proposition 3.3. IfX is an almost Lindel¨of space(a weakly Lindel¨of space), then every clopen subset ofX is almost Lindel¨of(respectively, weakly Lindel¨of).
Proposition 3.4. The sumL
s∈SXs is almost Lindel¨of (weakly Lindel¨of)if and only if all spacesXsare almost Lindel¨of(respectively, weakly Lindel¨of)and the setS is countable.
Proposition 3.5. A continuous image of an almost Lindel¨of space(a weakly Lindel¨of space)is almost Lindel¨of(respectively, weakly Lindel¨of).
Next, we turn to consider preimages. To show that the preimage of an almost Lindel¨of space (a weakly Lindel¨of space) under a closed 2-to-1 continuous map need not be almost Lindel¨of(respectively, weakly Lindel¨of) we use the Alexandorff duplicateA(X) of a spaceX. The underlying set ofA(X) isX× {0,1}; each point ofX× {1}is isolated and a basic neighborhood of a pointhx,0i ∈X× {0}is of the from (U× {0})∪((U× {1})\ {hx,1i}), whereU is a neighborhood ofxinX. It is well known thatX is Lindel¨of if and only if A(X) is Lindel¨of. But the statement is not true for almost Lindel¨of Urysohn spaces and weakly Lindel¨of spaces.
Example 3.6. There exists a closed 2-to-1 continuous map f : A(X) → X such thatX is a Uryshon almost Lindel¨of space, butA(X) is not a weakly Lindel¨of space (hence is not almost Lindel¨of).
Proof. LetX be the space from Example 2.2. ThenX is almost Lindel¨of and has an infinite discrete closed subset A= {aα : α < ω1}. Hence the Alexandroff duplicateA(X) ofX is not weakly Lindel¨of, sinceA×{1}is an uncountable infinite discrete, open and closed set in A(X). Let f : A(X) → X be the natural map.
Thenf is a closed 2-to-1 continuous map, which completes the proof.
If in the previous argument we use Example 2.3 instead of Example 2.2, we get the following:
Example 3.7. There exists a closed 2-to-1 continuous map f :X →Y such thatY is a Tychonoff weakly Lindel¨of space, butX is not weakly Lindel¨of.
Remark 3.1. The proof of Example 3.6 shows that the Alexandorff duplicate A(X) need not be almost Lindel¨of for a Urysohn almost Lindel¨of space X and the proof of Example 3.7 shows that the Alexandorff duplicateA(X) need not be weakly Lindel¨of for a Tychonoff weakly Lindel¨of spaceX.
Proposition 3.8. For a space X, the following conditions are equivalent:
(1) X is Lindel¨of;
(2) A(X) is Lindel¨of;
(3) A(X) is almost Lindel¨of;
(4) A(X) is weakly Lindel¨of.
Proof. The implications (1) ⇒ (2) ⇒ (3) ⇒ (4) are obvious. To show that (4)⇒(1), suppose thatX is not Lindel¨of. LetU be an open cover ofX witnessing that X is not Lindel¨of. Then{U× {0,1}:U ∈ U}is an open cover ofA(X) that witnesses thatA(X) is not weakly Lindel¨of, since all points ofX× {1}are isolated inA(X), which completes the proof.
Recall from [7] that a mappingf from a spaceX to a spaceY is calledalmost openiff−1(U)⊆f−1(U) for each open subset U of Y.
Proposition 3.9. If f : X → Y is an almost open and perfect continuous mapping andY is an almost Lindel¨of space, thenX is almost Lindel¨of.
Proof. LetU be an open cover ofX. Then there is a finite subfamily Uy ofU such that
f−1(y)⊆[
Uy for eachy∈Y.
LetUy=S
Uy. ThenVy=Y \f(X\Uy) is an open neighborhood ofy, sincef is closed. Let V ={Vy :y ∈Y}, then V is an open cover ofY, hence there exists a countable subfamily{Vyn :n∈ω}ofV such thatY =S
n∈ωVyn, sinceY is almost Lindel¨of. Sincef is almost open, then
X =f−1([
n∈ω
Vyn) = [
n∈ω
f−1(Vyn)⊆ [
n∈ω
f−1(Vyn)
⊆ [
n∈ω
Uyn⊆ [
n∈ω
[Uyn⊆ [
n∈ω
[{U :U ∈ Uyn},
sinceUyn is finite. Hence X is almost Lindel¨of , which completes the proof.
Similar to the proof of Proposition 3.9, we can prove the following proposition.
Proposition 3.10. If f : X → Y is an almost open and perfect continuous mapping andY is a weakly Lindel¨of spaces, thenX is weakly Lindel¨of.
It is well known that the product of two Lindel¨of spaces need not be Lindel¨of, which shows that the product of two almost Lindel¨of need not be almost Lindel¨of, since every Lindel¨of space is almost Lindel¨of and every almost Lindel¨of space is Lindel¨of for regular spaces. Since the product of a Lindel¨of space and a compact space is Lindel¨of, then the product of a regular almost Lindel¨of space and a compact space is almost Lindel¨of. For almost Lindel¨of spaces, we have the similar result.
Proposition 3.11. If X is almost Lindel¨of and Y is a compact space, then X×Y is almost Lindel¨of.
Proof. LetU be an open cover of X×Y. Without loss of generality we can assume thatUconsists of basic open sets ofX×Y. Since{x}×Y is a compact subset ofX×Y for eachx∈X, there exists a finite subfamily{Uxi×Vxi:i= 1,2, ...nx} ofU such that
{x} ×Y ⊆[
{Uxi×Vxi : 1≤i≤nx}.
LetWx=T
{Uxi : 1≤i≤nx}. Then {x} ×Y ⊆[
{Wx×Vxi : 1≤i≤nx}.
LetW={Wx:x∈X}. ThenWis an open cover ofX. SinceXis almost Lindel¨of, there is a countable subfamily{Wxj :j∈ω}ofW such thatX=S
j∈ωWxj, since X is almost Lindel¨of. Let
V ={Uxj i×Vxj i: 1≤i≤nxj, j∈ω}.
ThenV is a countable subfamily ofU. To show thatX×Y =S
{O:O ∈ V}, let hs, ti ∈X×Y be fixed. LetUs×Vtbe any open neighborhoods ofhs, tiin X×Y whereUsandVtare open neighborhood ofxandy inX andY, respectively. Since X =S
j∈ωWxj, then there exists aj∈ω such thats∈Wxj. Thus (Us×Vt)∩([
{Wxj ×Vxj i: 1≤i≤nxj})6=∅.
Therefore
(Us×Vt)∩([
{Uxj i×Vxj i: 1≤i≤nxj})6=∅.
We have hs, ti ∈[
{Uxj i×Vxj i: 1≤i≤nxj}=[
{Uxj i×Vxj i: 1≤i≤nxj}.
This implies hs, ti ∈ S
{O:O ∈ V}. Hence X×Y =S
{O :O∈ V}, which shows thatX×Y is almost Lindel¨of.
Similar to the proof of Proposition 3.11, we can prove the following proposition.
Proposition 3.12. If X is weakly Lindel¨of andY is a compact space, then X×Y is weakly Lindel¨of.
Asknowlegement. The authors would like to thank Prof. R.Li for his helpful discussions. They would also like to thank the referees for their kind help and valuable suggestions.
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(received 11.03.2009, in revised form 17.04.2009)
Institute of Mathematics, School of Mathematics and Computer Sciences, Nanjing Normal Uni- versity, Nanjing 210097, P. R. China
E-mail:[email protected]
